written 5.0 years ago by | modified 2.1 years ago by |
$A=\begin{bmatrix} 2& 1 & 1 \ 0& 1& 0 \ 1& 1& 2\end{bmatrix}$
written 5.0 years ago by | modified 2.1 years ago by |
$A=\begin{bmatrix} 2& 1 & 1 \ 0& 1& 0 \ 1& 1& 2\end{bmatrix}$
written 5.0 years ago by | modified 2.1 years ago by |
Solution:
$A=\begin{bmatrix} 2 & 1 & 1 \\ 0 & 1 & 0 \\ 1 & 1 & 2 \end{bmatrix}$
$|A|=1$
Characteristic equation is $|A-\lambda I|=0$
$\begin{vmatrix} 1-\lambda & 2 & -2 \\ -1 & 3-\lambda & 0 \\ 0 & -2 & 1-\lambda \end{vmatrix}=0$
$\lambda^{3}-(5) \lambda^{2}+(3+3+1+2-0-0)\lambda-1=0$
$\lambda^{3}-5\lambda^{2}+9\lambda-1=0$ -------------(1)
Cayley-Hamilton Theorem states that this equation is satisfied by the matrix A.
$\therefore$ Equation (1) becomes $A^{3}-5A^{2}+9A-I=0$ -------------(2)
$A=\begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}$ ---------------(3)
$A^{2}=A.A=\begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}=\begin{bmatrix} -1 & 12 & -4 \\ -4 & 7 & 2 \\ 2 & -8 & 1 \end{bmatrix}$ -------------(4)
$A^{3}=A.A^{2}=\begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix} \begin{bmatrix} -1 & 12 & -4 \\ -4 & 7 & 2 \\ 2 & -8 & 1 \end{bmatrix}=\begin{bmatrix} -13 & 42 & -2 \\ -11 & 9 & 10 \\ 10 & -22 & -3 \end{bmatrix}$ -------------(5)
$I=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ ---------------(6)
Put (3), (4), (5), (6) in (2),
$\begin{bmatrix} -13 & 42 & -2 \\ -11 & 9 & 10 \\ 10 & -22 & -3 \end{bmatrix}-5\begin{bmatrix} -1 & 12 & -4 \\ -4 & 7 & 2 \\ 2 & -8 & 1 \end{bmatrix}+9\begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}-\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$
$\therefore$ Cayley-Hamilton Theorem verified.
To find $A^{-1} \Longrightarrow$
Multiply equation (2) by $A^{-1}$
$(A^{3})A^{-1}-5A^{2}(A^{-1})+9A(A^{-1})-I(A^{-1})=0$
$A^{2}-5A+9I-A^{-1}=0$ ----------------(7)
Put equation (3), (4), (6) in (7),
$A^{-1}=\begin{bmatrix} -1 & 12 & -4 \\ -4 & 7 & 2 \\ 2 & -8 & 1 \end{bmatrix} -5 \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}+9\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}$